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Which of the following orbital designations is(are) not possible?
$\begin{array}{ll}{\text { a. } 3 f} & {\text { c. } 2 d} \\ {\text { b. } 4 d} & {\text { d. } 1 p}\end{array}$
So if you have an orbital in the p sub shell, L is going to be equal to one, and if you have an orbital in the G sub shell, then L is going to be equal to four.
Hello students in this question using s. Com api carmody and notations we have to describe the orbital in which following quantum numbers are present. So A which is given as an equals to one and L equals to zero. So this is the principal quantum number and this is the estimate quantum number. So principal quantum number is one. So we get one and four L equals to zero. We have the orbital which is S So this becomes the answer for the part is okay. Now, for the B part in which we have N equals 23 and L equals to one and for the N equals 23 We have three and for the L equals to one we have orbital which is the peak. So this Configuration is Envy three d. Okay? No, for the see part in which we have given N equals to four and L equals two. So for the fourth quantum number we have four. And for the eloquence to to the orbital is the D. Okay, So we get from here for so this become the answer for the part C. Okay, now, for the party in which we have N equals to four and L it is equal to three, so N equals to four is fourth. Okay. And equals to three. It is for the for the F. Okay, So this is the four F. Which is the orbital for this given configuration. So these are the different answers for this question. Okay, Thank you.
A useful tool to have one. Writing electron configurations is an outboard diagram. This tells us the order in which the sub levels air filled by electrons brilliant has a total of for electron. First, the one s orbital is filled up by the 1st 2 electrons. You can write that want us to then the next to go into the two s orbital to us too. Pensions? That's all four electrons. We start there. Aluminum has 13 ultra since its atomic number 13. It starts by killing the one s horrible with two electrons. Then the two s orbital, two electrons, then the two peeing orbital with six electrons in the three s with two electron. Now this accounts for 12 of the 13 electrons and so our last electron will go into the three p. Nitrogen is atomic number is seven. So what have seven electrons again? It's first to going to the oneness orbital, then two more into the two I sorbitol and discounts for four of the seven electrons. So the last three were put into the to P orbital Sodium has 11 electrons. It starts by feeling in the one s orbital ago with two elections comes into the to us or girl. Then it feels the to P orbital completely with six electrons, and this accounts for 10 of the 11 elect trends on our last one is put into the three Ask orbital.
Trying to figure out which of the following our middle designations are inaccurate. In order to do this, they need to know the relationship between and the principal energy level and its cub livers. And the principal energy level is a number that there's us. How many sub levels are present? For example, when N is equal to one, the number of sub lovers present are one. And this is gonna be an s orbited men and is equal to two. The number of sub levels present are too. And it will be to us arbiter as well as a true PR bitter. When N is equal to tree again, we'll have three sub lovers three s, 23 p and three d the same Damon and is equal to full. They have four sub lovers. Sorry. Pretend that has to be a full four p for duty. And so if hands did he f is not the possible designation. Two d is not a possible does ignition and Mont B is not a possible resignation. Only four D as possible